E=KE+PE, x,v,a calc question

[tony873004]

This problem is from the 2009 AP C Mechanics Exam. I get the same answers as the posted solutions except for part d. I'm hoping someone can tell me what I'm doing wrong. I'm going to show my work for part a and part c even though I got them correct, because they are used for part d.

A 3.0 kg object is moving along the x-axis in a region where its potential energy as a function of x is given as , where U is in joules and x is in meters. When the object passes the point x = -0.50 m, its velocity is + 2.0 m/s. All forces acting on the object are conservative.

(a) Calculate the total mechanical energy of the object.
$$E = KE + PE = \frac{1}{2}mv^2 + 4x^2 \\ = \frac{1}{2}\left( {3\,{\rm{kg}}} \right)\left( {2\,{\rm{m/s}}} \right)^2 + 4\left( { - 0.5\,{\rm{m}}} \right)^2 =7 J \\$$
(b) Calculate the x-coordinate of any points at which the object has zero kinetic energy.
(c) Calculate the magnitude of the momentum of the object at x = 0.60 m.
Come up with a formula for velocity as a function of time, using the formula derived in part a. This velocity formula will also be used in part d.
$$KE = \frac{1}{2}mv^2 = E - U = 7 - 4x^2 \\ v = \sqrt {\frac{{14 - 8x^2 }}{m}} = \sqrt {\frac{{14 - 8\left( {0.6} \right)^2 }}{3}} = 1.93\,{\rm{m/s}} \\ p = mv = \left( 3 \right)\left( {1.93} \right) = 5.77\,{\rm{kg - m/s}} \\$$

(d) Calculate the magnitude of the acceleration of the object as it passes x = 0.60 m.
acceleration is the derivative of velocity, so take the derivative and set t to 0.6.
$$KE\left( x \right) = 7 - U\left( x \right) = 7 - 4x^2 \\ v\left( x \right) = \sqrt {\frac{{2\left( {7 - 4x^2 } \right)}}{m}} \\ a\left( x \right) = \frac{1}{2}\frac{2}{m}\left( { - 8x} \right)\left[ {\frac{{2\left( {7 - 4x^2 } \right)}}{m}} \right]^{ - 1/2} \\ \frac{{ - 8x}}{m}\left[ {\frac{{2\left( {7 - 4x^2 } \right)}}{m}} \right]^{ - 1/2} \\ \frac{{ - 8\left( {0.6} \right)}}{3}\left[ {\frac{{2\left( {7 - 4\left( {0.6} \right)^2 } \right)}}{3}} \right]^{ - 1/2} =-0.83 m/s^{2} \\$$

But this is not the same answer as the scoring guide. They get 1.6 m/s².
I know my velocity formula is correct, because I got the correct momentum in part c. I'm pretty sure I took the derivative correctly. I even verified that with graphing software.

Additionally, I did a double check by using my velocity formula to compute the velocity at 0.6 s and 0.6001 s. The acceleration would then be (v₂-v₁)/Δx , where Δx is 0.0001. I also get -0.83 m/s² with this method.

How did I get the same wrong answer using 2 different methods?

Their justification is here: http://apcentral.collegeboard.com/apc/public/repository/ap09_physics_c_mechanics_sgs.pdf

Thanks!

 

[cepheid]

Yeah, tricky. I think you're the one who's in the wrong though, unfortunately. You solved part d by assuming this:

$a(x) = \frac{d}{dx}v(x)$ × WRONG

What's actually true is that acceleration is the TIME derivative of velocity:$$a(x) = \frac{d}{dt}v(x)$$ and so you have to differentiate using the chain rule, just like they did in the solutions manual.

Your first method was to calculate dv/dx exactly, and your second method was to calculate Δv/Δx for a small but finite Δx of 0.0001. Both of these were wrong, because neither of them is the acceleration, which is dv/dt (instantaneous) or Δv/Δt (average).

This problem is from the 2009 AP C Mechanics Exam. I get the same answers as the posted solutions except for part d. I'm hoping someone can tell me what I'm doing wrong. I'm going to show my work for part a and part c even though I got them correct, because they are used for part d.

A 3.0 kg object is moving along the x-axis in a region where its potential energy as a function of x is given as , where U is in joules and x is in meters. When the object passes the point x = -0.50 m, its velocity is + 2.0 m/s. All forces acting on the object are conservative.

(a) Calculate the total mechanical energy of the object.
$$E = KE + PE = \frac{1}{2}mv^2 + 4x^2 \\ = \frac{1}{2}\left( {3\,{\rm{kg}}} \right)\left( {2\,{\rm{m/s}}} \right)^2 + 4\left( { - 0.5\,{\rm{m}}} \right)^2 =7 J \\$$
(b) Calculate the x-coordinate of any points at which the object has zero kinetic energy.
(c) Calculate the magnitude of the momentum of the object at x = 0.60 m.
Come up with a formula for velocity as a function of time, using the formula derived in part a. This velocity formula will also be used in part d.
$$KE = \frac{1}{2}mv^2 = E - U = 7 - 4x^2 \\ v = \sqrt {\frac{{14 - 8x^2 }}{m}} = \sqrt {\frac{{14 - 8\left( {0.6} \right)^2 }}{3}} = 1.93\,{\rm{m/s}} \\ p = mv = \left( 3 \right)\left( {1.93} \right) = 5.77\,{\rm{kg - m/s}} \\$$

(d) Calculate the magnitude of the acceleration of the object as it passes x = 0.60 m.
acceleration is the derivative of velocity, so take the derivative and set t to 0.6.
$$KE\left( x \right) = 7 - U\left( x \right) = 7 - 4x^2 \\ v\left( x \right) = \sqrt {\frac{{2\left( {7 - 4x^2 } \right)}}{m}} \\ a\left( x \right) = \frac{1}{2}\frac{2}{m}\left( { - 8x} \right)\left[ {\frac{{2\left( {7 - 4x^2 } \right)}}{m}} \right]^{ - 1/2} \\ \frac{{ - 8x}}{m}\left[ {\frac{{2\left( {7 - 4x^2 } \right)}}{m}} \right]^{ - 1/2} \\ \frac{{ - 8\left( {0.6} \right)}}{3}\left[ {\frac{{2\left( {7 - 4\left( {0.6} \right)^2 } \right)}}{3}} \right]^{ - 1/2} =-0.83 m/s^{2} \\$$

But this is not the same answer as the scoring guide. They get 1.6 m/s².
I know my velocity formula is correct, because I got the correct momentum in part c. I'm pretty sure I took the derivative correctly. I even verified that with graphing software.

Additionally, I did a double check by using my velocity formula to compute the velocity at 0.6 s and 0.6001 s. The acceleration would then be (v₂-v₁)/Δx , where Δx is 0.0001. I also get -0.83 m/s² with this method.

How did I get the same wrong answer using 2 different methods?

Their justification is here: http://apcentral.collegeboard.com/apc/public/repository/ap09_physics_c_mechanics_sgs.pdf

Thanks!

 

[Doc Al]

Redo your derivative. Looks like you messed up when applying the chain rule. And as cepheid explains, you need the derivative with respect to time to do it your way.

The easy way to find the acceleration is to first find the force: F = - dU/dx

 

[tony873004]

I was pretty sure I was the one who got it wrong. If they made a mistake, they would have noticed it after grading a few thousand exams! Thanks Cephid and Doc Al. I could have stared at that for another hour and not noticed my wrong variable!

 

[Nickie]

I can understand your frustration and confusion with getting the same wrong answer using two different methods. However, it is important to remember that there is always a possibility of making mistakes in calculations, even for experts. In this case, it is possible that there is a small error in your calculations that is causing the difference in the answer.

To help you find the error, I would recommend going through your calculations step by step and double checking each step. It might also be helpful to have a colleague or a tutor review your work to see if they spot any mistakes.

Another possibility is that there could be a typo or an error in the scoring guide. While this is less likely, it is still worth considering and double checking all the numbers and formulas used in the scoring guide as well.

In any scientific problem, it is important to always double check and verify your results to ensure accuracy. Keep in mind that making mistakes is a natural part of the scientific process and it is important to learn from them and improve in the future. I hope this helps and good luck with your calculations!